# Proving $[(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})]\cdot(\vec{c}+\vec{a})=2\vec{c}\cdot(\vec{b}\times\vec{a})$

$$[(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})]\cdot(\vec{c}+\vec{a})=2\vec{c}\cdot(\vec{b}\times\vec{a})$$

I'm supposed to prove that this is true for all vectors $a,b,c$, but I keep getting $4(\vec{a}\times \vec{b})\cdot \vec{c}=0$, which is obviously not true.

Seems to me like there is a mistake in the problem, that it should be $\vec{a}\times\vec{b}$ instead of $\vec{b}\times\vec{a}$. Or I'm doing it wrong?

I'm using facts that $\vec{a}\times\vec{a}=0$ and $[\vec{a},\vec{b},\vec{c}]=[\vec{c},\vec{b},\vec{a}]=-[\vec{b},\vec{a},\vec{c}]$.

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That's what I got too: using the fact that $$\vec{b}\times \vec{b}=\vec{0}$$ and $$(\vec{b}\times \vec{c})\cdot\vec{c}=0$$ we can expand the left hand side as follows: $$[(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})]\cdot(\vec{c}+\vec{a})$$ $$=[\vec{a}\times\vec{b}+\vec{a}\times\vec{c}+\vec{b}\times \vec{c}]\cdot(\vec{c}+\vec{a})$$ $$=[\vec{a}\times\vec{b}+\vec{a}\times\vec{c}+\vec{b}\times \vec{c}]\cdot(\vec{c}+\vec{a})$$ $$=(\vec{a}\times \vec{b})\cdot\vec{c}+(\vec{b}\times \vec{c})\cdot\vec{a}$$ $$=2\vec{c}\cdot(\vec{a}\times\vec{b})$$ where the last equality follows from $$(\vec{b}\times \vec{c})\cdot\vec{a}=\vec{a}\cdot(\vec{b}\times \vec{c})=(\vec{a}\times \vec{b})\cdot\vec{c}.$$